Question regarding the extreme wind velocity distribution in Eurocode

Question

Above is the definition of the probability factor from Eurocode 1-4. It is used in context of calculating the extreme wind load on a building. Ignoring some other factors, it is used as follows:

$$v_b = v_{b,0} * c_{prob}$$

where $v_{b,0}$ is the extreme wind velocity with less than 0.02 probability of being exceeded. The factor $c_{prob}$ is used to reduce the extreme wind velocity in cases when considering time periods shorter than the 50 year-return period (0.02 probability).

I assume the formula comes from Gumbel distribution:

$$p = e^{-e^{-\frac{v-\mu}{\beta}}}$$

which can be solved for $v$:

$$v = \mu-\beta \ln(-\ln(p))$$

where $p$ is the probability of encountering extreme wind $v$.

Taking the ratio $\frac{v}{v_{b,0}}$ and considering that $v_{b,0}$ is defined to have to probability of being exceeded of 0.02, we get:

$$\frac{v}{v_{b,0}}=\frac{\mu-\beta \ln(-\ln(1-p))}{\mu-\beta \ln(-\ln(0.98))}$$

Defining $K = \frac{\beta}{\mu}$, we get:

$$\frac{v}{v_{b,0}} = c_{prob} = \frac{1-K \ln(-\ln(1-p))}{1-K \ln(-\ln(0.98))}$$

Finally to my question: Where does the exponent n come from in the Eurocode formula? As I have derived to formula in the picture, there is no need for the exponent. Is my derivation flawed? Is there something I don't know?

EDIT: Only reason I assumed Gumbel distribution was that I was able to get a similar formula using it. There seems to be nothing in Eurocodes that goes into detail as to what distribution is actually used.

Answer 1

Your derivation seems absolutely reasonable, as does your assumption about the Gumbel Distribution, with one minor detail.

IMHO, the reason why the exponent $n$ exists, is that the Eurocode does not really care about the velocity but about wind pressures (ultimately about loads on the structure). So the idea, is that you are trying to capture the extreme value of the wind pressure (though the wind speed).

So all the analysis you carried out is not about $\frac{v}{v_{b,0}}$ but about wind pressures $\frac{q}{q_{b,0}}$. So, instead its:

$$\frac{q}{q_{b,0}}= c_{\color{red}{q},prob} = \frac{1-K \ln(-\ln(1-p))}{1-K \ln(-\ln(0.98))} $$

However, because the wind pressure $q$ is proportional to the square of the wind velocity so:

$$\frac{q}{q_{b,0}} = \frac{v^2}{v_{b,0}^2}=\left(\frac{v}{v_{b,0}}\right)^2 $$

$$\left(\frac{q}{q_{b,0}} \right)^{0.5} = \frac{v}{v_{b,0}}$$

$$\left(c_{q,prob} \right)^{0.5} = \frac{v}{v_{b,0}}$$ $$\left(\frac{1-K \ln(-\ln(1-p))}{1-K \ln(-\ln(0.98))} \right)^{0.5} = \frac{v}{v_{b,0}}$$

However, because $c_{prob}$ is a factor that is applied to velocity:

$$ \frac{v}{v_{b,0}} = \left(\frac{1-K \ln(-\ln(1-p))}{1-K \ln(-\ln(0.98))} \right)^{0.5}$$